Difference between revisions of "2009 IMO Problems/Problem 2"
Qwertysri987 (talk | contribs) (→Solution) |
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== Solution == | == Solution == | ||
− | + | ===Diagram=== | |
+ | <asy> | ||
+ | dot("O", (50, 43), W); | ||
+ | dot("A", (40, 100), N); | ||
+ | dot("B", (0, 0), S); | ||
+ | dot("C", (100, 0), S); | ||
+ | dot("Q", (24, 60), W); | ||
+ | dot("P", (52, 80), E); | ||
+ | dot("L", (62, 30), SE); | ||
+ | dot("M", (38, 70), N); | ||
+ | dot("K", (26, 40), S); | ||
+ | draw((100, 0)--(24, 60), dotted); | ||
+ | draw((0, 0)--(52, 80), dashed); | ||
+ | draw((0, 0)--(100, 0)--(40, 100)--cycle); | ||
+ | draw((24, 60)--(52, 80)); | ||
+ | draw((26, 40)--(38, 70)--(62, 30)--cycle); | ||
+ | draw(circle((47, 48), 23)); | ||
+ | </asy> | ||
+ | Diagram by qwertysri987 | ||
+ | ---------------------------- | ||
By parallel lines and the tangency condition, <cmath>\angle APM\cong \angle LMP \cong \angle LKM.</cmath> Similarly, <cmath>\angle AQP\cong \angle KLM,</cmath> so AA similarity implies <cmath>\triangle APQ\sim \triangle MKL.</cmath> Let <math>\omega</math> denote the circumcircle of <math>\triangle ABC,</math> and <math>R</math> its circumradius. As both <math>P</math> and <math>Q</math> are inside <math>\omega,</math> | By parallel lines and the tangency condition, <cmath>\angle APM\cong \angle LMP \cong \angle LKM.</cmath> Similarly, <cmath>\angle AQP\cong \angle KLM,</cmath> so AA similarity implies <cmath>\triangle APQ\sim \triangle MKL.</cmath> Let <math>\omega</math> denote the circumcircle of <math>\triangle ABC,</math> and <math>R</math> its circumradius. As both <math>P</math> and <math>Q</math> are inside <math>\omega,</math> | ||
Revision as of 00:18, 22 July 2020
Problem
Let be a triangle with circumcentre
. The points
and
are interior points of the sides
and
respectively. Let
and
be the midpoints of the segments
and
, respectively, and let
be the circle passing through
and
. Suppose that the line
is tangent to the circle
. Prove that
.
Author: Sergei Berlov, Russia
Solution
Diagram
Diagram by qwertysri987
By parallel lines and the tangency condition, Similarly,
so AA similarity implies
Let
denote the circumcircle of
and
its circumradius. As both
and
are inside
It follows that